Number System Tutorial II: Divisibility, Remainder, HCF & LCM

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Remainders

There are a plethora of remainder related questions you can expect in your exam. So it is a mandatory requirement to create a clear awareness of remainders before going to attend the exam.

Based on above illustration, any number 'N' can be expressed as `N = dq + r`.

Properties of remainders

  • When the dividend is less than divisor, then the remainder is the dividend itself. Eg: when 5 divided by 7, the remainder is 5 itself, and then the quotient of this division is 0.
  • Remainders are consistent under basic mathematical operations such as addition, subtraction, multiplication and powers.
    Addition:When `N_1` and `N_2` (two distinct dividends) divided by a divisor `d`, leaves the respective remainders `r_1` and `r_2`, then the remainder when `N_1 + N_2` divided by `d` is `r_1 + r_2`.
    Eg. `"Rem"[11/9] = 2` and `"Rem"[14/9] = 5`
    `:. "Rem"[(11 + 14)/9] = 2 + 5 = 7`
    Subtraction:When `N_1` and `N_2` (two distinct dividends) divided by a divisor `d`, leaves the respective remainders `r_1` and `r_2`, then the remainder when `N_1 - N_2` divided by `d` is `r_1 - r_2`.
    Eg. `"Rem"[19/10] = 9` and `"Rem"[15/10] = 5`
    `:. "Rem"[(19 - 15)/10] = 9 - 5 = 4`
    Multiplication:When `N_1` and `N_2` (two distinct dividends) divided by a divisor `d`, leaves the respective remainders `r_1` and `r_2`, then the remainder when `N_1 xx N_2` divided by `d` is `r_1 xx r_2`.
    Eg. `"Rem"[14/10] = 4` and `"Rem"[12/10] = 2`
    `:. "Rem"[(14 xx 12)/10] = 4 xx 2 = 8`
    Powers or exponents:When a dividend 'N' divided by divisor 'd', leaves a remainder 'r', then `N^n` leaves a remainder `r^n` when it is divided by d.
    Eg. `"Rem"[10/8] = 2`
    `:. "Rem"[10^2/8] = 2^2 = 4`
  • Concept of Negative remainders: As per Eucledian Lemma any number 'N' can be expressed in the following way; `N = dq + r`
    In a similar arrangement, 7 can be expresses as; `7 = 3 × 2 + 1`, where 3 is the divisor, 2 is the quotient and 1 is the remainder.
    7 can be expressed in the following way too; `7 = 3 × 3 - 2`, where 3 is the same divisor which we considered above, 3 is the new quotient and '-2' is the remainder as per the change in the quotient. Therefore the remainders 1 and -2 are representing the same division. Hence -2 is the corresponding negative remainder of the original remainder 1, when the divisor is 3.
    `"Negative remainder" = "original remainder" - "divisor"`
    `"Therefore original remainder" = "divisor" + "negative remainder"`
    Let's look at one more example to understand concept of negative remainders.
    `"Rem"[18/8] = 2` and `"Rem"[5/8] = 5`
    `:. "Rem"[(18-5)/8] = 2 - 5 = -3`
    `"Hence the original remainder" = "divisor" + "negative remainder"`
    `= 8 + (-3) = 5`
    ie; `"Rem"[(18-5)/8] = "Rem"[13/8] = 5`
  • Pattern of remainders:

    Pattern of remainders is one of the most aesthetical and very important features of remainders. Let us consider the following pattern of remainders;

    `"Rem"[3^1/5] = 3`
    `"Rem"[3^2/5] = 3 xx 3 -> 4`
    `"Rem"[3^3/5] = 4 xx 3 -> 2`
    `"Rem"[3^4/5] = 2 xx 3 -> 1`
    `"Rem"[3^5/5] = 1 xx 3 -> 3` and so on...

    The above pattern of remainders will repeat cyclically after the occurrence of 1 and the pattern of remainders is 3,4,2,1, 3,4,2,1... In the given example, there are 4 distinct remainders. If the exponent of 3 is any multiple of 4, then the remainder should be 1; Ie. if the exponent is in the form `4k + 1 -> "remainder is 3"`
    `4k + 2 -> "remainder is 4"`
    `4k + 3 -> "remainder is 2"`
    `4k + 0 -> "remainder is 1"`, Where k is any whole number, k = 0, 1, 2, 3...

    Find the remainder when `3^1729` divided by 5
    As per the above explanation, there are 4 distinct remainders and those are occurring in the order 3, 4, 2, 1. `1729 = 4k + 1` (when 1729 divided by 4, it leaves the remainder 1). If the exponent is in the form `4k + 1`, then the remainder is 3.
    Find the remainder when `13^271` divided by 7
    First of all develop the cyclic pattern of possible remainders.
    `"Rem"[13^1/7] = 6`
    `"Rem"[13^2/7] = 6 xx 6 -> 1`
    `"Rem"[13^3/7] = 1 xx 6 -> 6`
    `"Rem"[13^4/7] = 6 xx 6 -> 1`
    Now we got a pattern of remainders in the cyclic order 6,1,6,1...
    Note: In the pattern of remainders if in any step the remainder occurs as 1, then the pattern will start from there. Hence the operation of finding remainders can stop at that point.
    In the above pattern, there are two distinct remainder, ie. when the exponent is odd, remainder is 6 and when the exponent is even, then the remainder is 1. `:. "Rem"[13^271/7] = 6`, because 271 is an odd exponent.

Fermat's little theorem

Let N and p are two relatively prime numbers and p itself is a prime, then `N^(p-1)/p` leaves a remainder of 1.

Eg. `"Rem"[100^36/37] = 1`

Find the remainder when `100^41` divided by 41
As per Fermat's theorem, `N^(p-1)/p` leaves a remainder of 1, where N and p are two relatively prime numbers and p itself is a prime. In the given question, the divisor 41 is a prime number and 100 and 41 are relatively prime.`:. "Rem"[100^40/41] = 1`.
So, `"Rem"[100^41/41] = "Rem"[(100^40 xx 100)/41] = "Rem"[100/41] = 18`

Wilson's theorem

For all prime `p`, `[(p - 1)! + 1]` is always divisible by `p`.
OR
When `(p-1)!` Is divided by `p`, the remainder is `(p - 1)`.

Let's look at an example to understand Wilson's theorem.

Find the remainder when 6! Divided by 7.
As per Wilson's theorem; 7 is a prime `-> "Rem"[((7-1)!)/7] = 7-1 = 6`

Remainder theorem and its numerical application

If `p(x)` is a polynomial in x, then `p(a)` is the remainder when `p(x)` divided by `(x - a)`.

Understanding the numerical application of this theorem will help you to solve some complicated type of remainder related questions. Let's take a look at few examples to understand remainder theorem better.

Find the remainder when `2^402` divided by 7.
Consider `2^402` as a polynomial in 2. Also, 7 can be expressed in terms of 2 as: `7 = 2^3 - 1`, which can be considered as a polynomial on `2^3`
`:. "Rem"[2^402/7] = "Rem"[(2^3)^134/ (2^3 - 1)] = 1^134 = 1`
Find the remainder when `16^133` divided by 17.
`17 = 16 + 1 -> ` it is in the form of `x + 1`, where `x = 16`
`16^133` is in the form of `x^133`, where `x = 16`
`"Rem"[16^133/(16 + 1)] = (-1)^133 = -1
Here the remainde is negative, so original remainder (positive) is 17 - 1 = 16
Find the remainder when `27^101` divided by 82.
`82 = 3^4 + 1`
`27^101 = (3^3)^101 = (3^4)^75 * 3^3 = (3^4)^75 * 27`
`"Rem"[27^101/82] = "Rem"[((3^4)^75 * 27)/(3^4 + 1)] = (-1)^75 * 27 = -27`
Here we got negative remainder as -27, so actual remainder is 82 - 27 = 55
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